What is the slope of the line tangent to $f(x) = -x^{2}+2x+4$ at $x = 1$ ?
Explanation: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-(x+\Delta x)^{2}+2(x+\Delta x)+4) - (-x^{2}+2x+4)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-(x^{2}+2x \Delta x+\Delta x^{2})+2(x+\Delta x)+4) - (-x^{2}+2x+4)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-x^{2}-2(x \Delta x)-\Delta x^{2}+2x+2(\Delta x)+4+x^{2}-2x-4}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-2(x \Delta x)-\Delta x^{2}+2(\Delta x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} -2x-\Delta x+2$ $ = -2x+2$ $ = (-2)(1)+2$ $ = 0$